In $\Bbb R^2,$ let $L$ be the line $y = mx.$ Find an expression for $T(x, y),$ where $T$ is the projection on $L$ along the line perpendicular to $L.$

Question

In $\Bbb R^2,$ let $L$ be the line $y = mx$, where $m\neq 0$. Find an expression for $T(x, y),$ where $T$ is the projection on $L$ along the line perpendicular to $L.$ (See the definition of projection in the exercises of Section $2.1.$)

This was a problem given in the book, "Linear Algebra" by Stephen H Friedberg, Insel and Spence on Chapter-2 (Section 2.5), page number-117 (4th Edition).

The definition of projection in the excercises of Section- $2.1$ is given as:

Let $V$ be a vector space and $W_1$ and $W_2$ be subspaces of $V$ such that $V = W1 ⊕ W2.$ ($\oplus$ stands for direct sum.) A function $T: V → V$ is called the projection on $W_1$ along $W_2$ if, for $x = x_1 + x_2$ with $x_1 ∈ W_1$ and $x_2 ∈ W_2,$ we have $T(x) = x_1.$

Complying with this definition, I considered the subspace of $\Bbb R^2$ consisting of all the points in line $L$ and the subspace consisting of all the points lying on the line perpendicular to $L$ and let us call it, $W_1$ and $W_2$ respectively

Now, the point is, there are infinite number of lines perpendicular to a fixed line. In our case, $\{y=-\frac 1m(x)+c:c\in \mathbb{R}\}$ are the set of lines perpendicular to $L$. I don't quite get out of all these line, which one are referring to in the question (, by the phrase, "along the line perpendicular to $L$ ").

So, I just chose an arbitrary lperpendicular to $L$ say, $P:y+\frac 1m(x)-c'=0$ and call the set of points on $P$ as $W_2.$

But then intuitively, it becomes very much obvious that $\Bbb R^2\neq W_1\oplus W_2,$ is the case, geometrically.

Due to this, the objective of the problem i.e, to find "an expression for $T(x, y),$ where $T$ is the projection on $L$ along the line perpendicular to $L$" makes no sense for the notion of a a projection of a subspace along another is valid only if the direct sum of those two subspaces equals the given vector space.

Am I missing something? Any help regarding this apparent issue will be greatly appreciated.

Answer 1

Well I think that as is written you don't have to consider one perpendicular line but every perpendicular line, let me explain.

I will deliberately choose $m=1$ this could be generalized easily. We have the line in the cartesian plane $y = x$ so $W_1 = \{ (x,y) \in \Bbb R^2 : x-y = 0\}$ and $W_2 = W_1^c$. How the projection needs to work? We want a projection on the respective perpendicular line so for $P\in W_2$, and we take the intersection of $y =x$ and perpendicular $(y-y_p) = -(x-x_p)$, the results is certainly a point in $W_1$ and we have our projection!

Answer 2

Let $L$ be the line $y = m x$.

The equation $L_1$ of the line perpendicular to $L$ and passing through a point $(a, b)$ is given by $$ y - b = -{1 \over m} (x - a) $$

When $L_1$ and $L$ intersect at a point $(x, y)$, both equations of the lines $L$ and $L_1$ are satisfied. Hence, we have $$ y = m x $$ $$ y - b = -{1 \over m} (x - a) $$

Solving these two equations, we get $$ x = {b + {a \over m} \over m + {1 \over m}}, \ \ y = {m (b + {a \over m}) \over m + {1 \over m}} $$

Finally, replacing $a$ and $b$ by $x$ and $y$, we get $$ T(x, y) = \left( {y + {x \over m} \over m + {1 \over m}}, {m (y + {x \over m}) \over m + {1 \over m}} \right) $$